# What is the equation of the normal line of f(x)= tan(2x)/cotx at x = pi/8?

##### 1 Answer
Sep 23, 2016

$- 2 \left(\sqrt{2} + 1\right) \left(2 - \sqrt{2}\right) \left(y - \frac{2 - \sqrt{2}}{2}\right) = x - \frac{\pi}{8}$

#### Explanation:

equation of the normal: $- y {'}_{0} \left(y - {y}_{0}\right) = x - {x}_{0}$

$y ' = \frac{\left(\frac{2}{\cos} ^ 2 \left(2 x\right)\right) \cdot \cot x + \frac{\tan 2 x}{\sin} ^ 2 x}{\cot} ^ 2 x$

$y {'}_{0} = \frac{\left(\frac{2}{\cos} ^ 2 \left(\frac{\pi}{4}\right)\right) \cdot \cot \left(\frac{\pi}{8}\right) + \frac{\tan \left(\frac{\pi}{4}\right)}{\sin} ^ 2 \left(\frac{\pi}{8}\right)}{\cot} ^ 2 \left(\frac{\pi}{8}\right) =$

$= \frac{4 \cos \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) + 1}{\cos} ^ 2 \left(\frac{\pi}{8}\right) =$

$= \frac{2 \sin \left(\frac{\pi}{4}\right) + 1}{\frac{\cos \left(\frac{\pi}{4}\right) + 1}{2}} =$

$= 4 \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 2}$

${y}_{0} = \tan \left(\frac{\pi}{8}\right) = 1 - \frac{\sqrt{2}}{2}$

$- 2 \left(\sqrt{2} + 1\right) \left(2 - \sqrt{2}\right) \left(y - \frac{2 - \sqrt{2}}{2}\right) = x - \frac{\pi}{8}$